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Part 5. Suppose that your hash function resolves collisions using the open addressing method with double hashing. The double

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Answer #1
Inserting 14
main hash value is 1
There is already an item in 1
double hash value is 4
1 + 1*4 = 5
So, checking at index 5 % 13 = 5
There is already an item in 5
double hash value is 4
1 + 2*4 = 9
So, checking at index 9 % 13 = 9
There is already an item in 9
double hash value is 4
1 + 3*4 = 13
So, checking at index 13 % 13 = 0
14 is inserted at position 0
14, 79, -, -, 17, 98, -, 59, -, 22, -, 37, -

HashTable
-----------
0   -      14
1   -      79
2   -
3   -
4   -      17
5   -      98
6   -
7   -      59
8   -
9   -      22
10      -
11      -      37
12      -

Answer:
--------
it's placed at index 0
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